Sine Function Period 92Cc3B

1. **State the problem:** We are given the function $$y=-2\sin\frac{\pi}{2}\left(x-\frac{1}{4}\right)$$ and need to find its period, interval, starting point, and 5 x-values (in radians) relevant to the sine wave. 2. **Recall the formula for the period of a sine function:** For a function $$y = A \sin(B(x - C))$$, the period $$T$$ is given by: $$T = \frac{2\pi}{|B|}$$ where $$B$$ is the coefficient of $$x$$ inside the sine. 3. **Identify $$B$$:** Here, $$B = \frac{\pi}{2}$$. 4. **Calculate the period:** $$T = \frac{2\pi}{\left|\frac{\pi}{2}\right|} = \frac{2\pi}{\frac{\pi}{2}} = 2\pi \times \frac{2}{\pi} = 4$$ So, the period is $$4$$. 5. **Determine the starting point:** The phase shift $$C$$ is $$\frac{1}{4}$$, so the graph starts at $$x = \frac{1}{4}$$. 6. **Determine the interval for one period:** Since the period is $$4$$ and the starting point is $$\frac{1}{4}$$, the interval for one full cycle is: $$\left[\frac{1}{4}, \frac{1}{4} + 4\right] = \left[\frac{1}{4}, \frac{17}{4}\right]$$ 7. **Find 5 x-values within one period:** We can choose 5 equally spaced points starting at $$\frac{1}{4}$$ with spacing $$\frac{T}{4} = 1$$: - $$x_1 = \frac{1}{4}$$ - $$x_2 = \frac{1}{4} + 1 = \frac{5}{4}$$ - $$x_3 = \frac{1}{4} + 2 = \frac{9}{4}$$ - $$x_4 = \frac{1}{4} + 3 = \frac{13}{4}$$ - $$x_5 = \frac{1}{4} + 4 = \frac{17}{4}$$ These are the 5 x-values in radians. **Final answers:** - Period: $$4$$ - Interval for one period: $$\left[\frac{1}{4}, \frac{17}{4}\right]$$ - Starting point: $$\frac{1}{4}$$ - 5 x-values: $$\frac{1}{4}, \frac{5}{4}, \frac{9}{4}, \frac{13}{4}, \frac{17}{4}$$